3 Icebergs in a bath

Dear Forum, I would appreciate if you could answer this question. (The question is taken from Thinking Physics by L.C.Epstein).

The question has 3 different baths, all full to the brim with varying icebergs floating in them, and it asks the reader what would happen to the water level in each of the baths if the icebergs melted.

The first has an iceberg with an airbubble in it.
The second has an iceberg with an unfrozen watterbubble in it.
The third has an iceberg with a solid railway spike in it.

An answer is also provided, but one section of it I find hard to grasp. The answer states that when the icebergs melt, the first and second baths would remain brim full, but the water level in the third bath would go down.

I understand the answer to the second and the third baths, but not the first.

The reason given for the first bath (iceberg+airbubble) remaining brim full is as follows. If you imagine moving the airbubble to the top of the berg, their would be no change in weight, and thus no change in displacement, hence the water level would not change. Then if you popped the airbubble, again no change in weight or displacement, hence no water level change. Then it is just a regular iceberg, and so when it melts it will put total water of equal mass into the bathtub and hence water level will remain the same.

This explanation seems reasonable to me, however it clashes with a different method of explanation which I have.

I happen to think that the water level would go down in the first bath, and heres why.

If you take the iceberg complete with airbubble, the total mass of the system is mass of berg+mass of bubble (air has mass too). Hence the total weight of system is weight of berg+weight of bubble. Hence weight of liquid displaced not only depends on the dislpacement to support the weight of berg, but also the displacement to support the weight of the bubble.Correct?

Now, when this melts, the weight of berg will transferred to the bathtub, but the weight of airbubble will float up into the atmosphere, hence the weight of water gained on the bathtub is due to the weight of the melted berg, but not weight of air, but the weight of air displaced some water in the first place.

So the brim full water level in the first place was due to the weight imparted by the berg+airbubble. But only the weight of berg is transferred to the bathtub, hence the water level must go down. The air just escapes from the water.

In other words, if the airbubble was replaced by a vacuum bubble then the water level would remain the same, as the vacuum bubble does not displace any water as it has no mass, so the displacement is entirely due to the berg, which melts and water level remains the same.

You're explanation would apply if the bubble were highly compressed and thus weighed more than an equal volume of ambient air. For an atmospheric-pressure bubble, think of what the result would be if you were to drill a vent hole through the top to the bubble. Through the bottom? Downward through the side below the waterline?

Thankyou for your reply, although it seems to be sending me in a similar direction to the way it is explained in the book. Which is what the problem is in the first place. Could you please further explain (or anyone else for that matter) why it makes a difference whether the weight of the bubble is above or below the weight of an equal volume of ambient air, as I see it any weight above zero would do as I explain.

In answer to your questions. Drilling to the hole from above, pressumably air from outside the berg would flow into the hole created by the drill, and the air in the bubble would remain in the bubble, also some water would be created as the drilled ice melted (the water would be equal in mass to the ice which had melted, so no net change in bathtub water level). Drilling to the hole from below, pressumably water would take up this newly created space, the melted ice from drilling would flow into the tank, and overall water level would decrease (the ice melted from drilling would create less volume of water than the newly created hole would occupy), so net decrease in water level in the bathtub. Unfortunately I dont understand the third question "Downward through the side below the waterline?" depending on where the airbubble is situated in the berg, I cant go downwards to the bubble, and below the waterline. But really I dont see the relevance of this explanation, after all im not going to be drilling holes, im just going to let it melt.

If you can, please try to explain in words rather than pictures or "imagine this......", "what if this......" statements.

Trapped air bubbles at the same density as the surrounding air reduce
the density of the iceberg.

As long as any trapped air bubbles are at the same density (pressure, temperature) as the surrounding air when they eventually get released by the melting iceberg, they have no effect on the water level.

If the icebergs were created at high pressure (2 kilo-bars or higher), they would have higher density (ice II or higher form), would sink, and the water level would rise as they melted.

Fortunately, on Earth the combination of low temperature and high pressures never occur naturally.

Thanks for the more detailed explanation, Jeff. I got too busy to respond again until now. Mr. Rich, you're analysis of the drilling effect is very impressive. Unfortunately, I didn't specify that I meant making the holes before introducing the berg to the water. The only relevant one is from the top, which was just to show that having a bubble inside the ice is the same as having an equal volume of ice missing from the top. The rest were just to get you thinking from all possible approaches. The one angled down below the waterline would let the air escape and be replaced by water. Since that water would come from the bathtub, it still wouldn't affect the level.

The only thing that puzzles me is why the water level in the third example would go down. In all of them, the iceberg and everything in it is already displacing its own weight. If the iceberg melts, it just becomes part of the water so it is no longer displacing its own weight of water but is filling that "void"- so the water level remains the same. Obviously unfrozen water will do the same. An air bubble reduces the weight displaced but also means there is less water in the iceberg. The iceberg with the railroad spike is heavier so displaces more water- but when the iceberg melts, the railroad spike is still there so-- OH! wait, the railroad spike doesn't float! its sinks to the bottom where it is supported by the bottom- it only displaces its own volume, not its own weight!

The only thing that puzzles me is why the water level in the third example would go down. In all of them, the iceberg and everything in it is already displacing its own weight. If the iceberg melts, it just becomes part of the water so it is no longer displacing its own weight of water but is filling that "void"- so the water level remains the same. Obviously unfrozen water will do the same. An air bubble reduces the weight displaced but also means there is less water in the iceberg. The iceberg with the railroad spike is heavier so displaces more water- but when the iceberg melts, the railroad spike is still there so-- OH! wait, the railroad spike doesn't float! its sinks to the bottom where it is supported by the bottom- it only displaces its own volume, not its own weight!

But as it sinks to the bottom, it punctures the container! ..or at the very least causes a disturbance of the total water volume at the surface, and thus water splashes over sides?

The melting sequences are pretty uneventfull for the first two scenarios, but the third scenario, the heavy-duty iron bar will fall to the bottom of the bath, whilst still having a considerable amount of ice encased around it, thus the volume/area of this falling through water,,causes a displacement action that disturbs the water in the bath.

Basically, the action of three induces a lot of water motion, whilst one and two are steady actions?